Question: $ \text{Find the mass of the surface } x^2 +y^2+z^2 = 16 \text{ where the density at a point is the distance to the xy-plane.} $

Distance from the xy plane can be given in terms of x and y is equal to z. So I solve for z and get $ z = \sqrt{16 - y^2 - x^2} $ I know that both y and x range from [0,4] to get me 1/8 of the sphere.

So the problem then becomes the following:

$ 8 \int_{0}^{4} \int_{0}^{4} \sqrt{16 - y^2 - x^2} dy dx$

But obviously that's a mess which makes me think I should be using spherical coordinates but I'm struggling to figure out how to work the density into that equation.

  • 2
    $\begingroup$ You have a two-dimensional integral; use polar coordinates. The integrand becomes $\sqrt{16-r^2} r \: dr \: d\theta$, integrated over a disc. $\endgroup$ – Michael Lugo Aug 14 '19 at 19:29
  • $\begingroup$ Surface has a thickness also? how does it vary with distance from $x-y$ plane? $\endgroup$ – Narasimham Aug 15 '19 at 3:41

The spherical integral for the surface mass is

$$I = \int_S |z|r^2\sin\theta d\theta d\phi$$

where $|z|$ is the density. Plug in the radius $r=4$ and the distance


and then carry out the integration

$$I=64 \int_0^{2\pi} d\phi \int_0^{\pi} \sin\theta|\cos\theta| d\theta = 128\pi$$

  • $\begingroup$ Would you mind elaborating a bit more on how you did what you did? |z| makes sense because that's the density, r^2 sin(theta) is a bit more shaky for me. Looks like x^2 + y^2 + z^2 = r^2 but then I'm not sure what the sin(theta) is doing. $\endgroup$ – financial_physician Aug 14 '19 at 20:40
  • $\begingroup$ Also I'm not sure how you got from the first equation to the second. r^3 = 64 which leaves just |cos(phi)|... integrated from [0,pi] would be another 2 to make it 128, but I'm not sure why the pi is coming from or if I'm thinking about it correctly $\endgroup$ – financial_physician Aug 14 '19 at 20:46
  • 1
    $\begingroup$ In spherical coordinates, $r^2\sin\theta d\theta d\phi$ is the standard representation of infinitesimal surface area, similar to $dx dy$ in the $xy$-coordinates. $\endgroup$ – Quanto Aug 14 '19 at 20:48
  • 1
    $\begingroup$ $2\pi$ comes from integration over $d\phi$ from the range $0$ to $2\pi$. Note that the density does not depend on $\phi$. $\endgroup$ – Quanto Aug 14 '19 at 20:51

First you want to find out the function $\vec{F}$. In this case, it is simply the z-component of the surface: $$\vec{F}=<0,0,z>$$ You are trying to compute the surface integral over a vector field: $$\int \int_S \vec{F} \cdot d\vec{S} = \int\int_{(u,v)\in D} \vec{F}(\vec\Phi) \cdot(\vec\Phi_u\times\vec\Phi_v)\ du\ dv$$ Next, you want to figure out how you want to parameterize the function. Since the surface is a sphere, it would stand to reason that you will want to use spherical coordinates. $$\vec\Phi(\phi,\theta)=<\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi>$$ $$\rho = \sqrt{16} = 4$$ $$0\leq\theta\leq 2\pi $$ $$0\leq\phi\leq \pi $$ So, $$\vec\Phi(\phi,\theta)=<4\sin\phi\cos\theta,4\sin\phi\sin\theta,4\cos\phi>$$ Take the partial derivatives in terms of $\phi$ and $\theta$. $$\vec\Phi_\phi(\phi,\theta)=<4\cos\phi\cos\theta,4\cos\phi\sin\theta,-4\sin\phi>$$ $$\vec\Phi_\theta(\phi,\theta)=<-4\sin\phi\sin\theta,4\sin\phi\cos\theta,0>$$ Then get the cross product of the two terms to get the normal vector: $$\vec\Phi_\phi(\phi,\theta)\times\vec\Phi_\theta(\phi,\theta) = <16\sin^2\phi\cos\theta, 16\sin^2\phi\sin\theta, 16\sin\phi\cos\phi> $$ And get the function with regards to $\Phi$: $$\vec{F}(\vec\Phi(\phi,\theta)) = <0,0,4\cos\phi> $$ Next you dot product the function and the normal vector: $$\vec{F}(\vec\Phi(\phi,\theta)) \cdot(\vec\Phi_\phi(\phi,\theta)\times\vec\Phi_\theta(\phi,\theta)) = 0+0+64\sin\phi\cos^2\phi$$ Then set up your integral: $$\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi} 64\sin\phi\cos^2\phi\ d\phi d\theta $$ $$\frac{128}{3}\int_{\theta=0}^{2\pi}d\theta = \frac{256\pi}{3}$$

Alternatively, since the surface is closed, you can utilize the Divergence Theorem (Gauss's Theorem) to calculate the mass. $$\int\int\int_E Div\vec{F}\ dV $$ $$Div\vec{F} = \vec{\nabla} \cdot \vec{F} = \left <\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right > \cdot \left <0,0,z \right > = 0+0+1 = 1$$ In this case: $$dV = \rho^2\sin\phi\ d\rho\ d\phi\ d\theta$$ and: $$\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi}\int_{\rho=0}^{4} \rho^2\sin\phi\ d\rho\ d\phi\ d\theta $$ $$=\frac{64}{3}\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi} \sin\phi\ d\phi\ d\theta$$ $$=\frac{128}{3}\int_{\theta=0}^{2\pi} d\theta=\frac{256\pi}{3}$$


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